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Phân tích đa thức thành nhân tử
(x+3)(x−6)+x2−9
Tk nha !
\(\left(x+3\right)\left(x-6\right)+x^2-9\)
\(=x^2-3x-18+x^2-9\)
\(=2x^2-3x-27\)
\(=\left(2x^2+6x\right)-\left(9x+27\right)\)
\(=\left(x+3\right)\left(2x-9\right)\)
Rút gọn thôi chứ phân tích sao được ._.
( x - 3 )2 - ( 4x + 5 )2 - 9( x + 1 )2 - 6( x - 3 )( x + 1 )
= x2 - 6x + 9 - ( 16x2 + 40x + 25 ) - 9( x2 + 2x + 1 ) - 6( x2 - 2x - 3 )
= x2 - 6x + 9 - 16x2 - 40x - 25 - 9x2 - 18x - 9 - 6x2 + 12x + 18
= -30x2 - 52x - 7
Sửa đề lại 1 chút là phân tích được mà bn Quỳnh:))
Ta có: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)
\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)
\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)
\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)
\(=\left(4x+7\right)\left(12x+17\right)\)
x^3 - 4x^2 + 4x + 4x - 8
= (X^3 - 8) - (4x^2 - 4x - 4x)
= (x - 2)(x^2 + 2x + 4) - 4x( x - 2)
= (x - 2)(x^2 + 2x + 4 - 4x)
= (x - 2)(x^2 - 2x + 4)
b) 4x^2 - 25 - (2x - 5)(2x- 7)
= (2x - 5)(2x + 5) - (2x - 5)(2x - 7)
= (2x - 5)(2x + 5 - 2x + 7)
= 12(2x - 5)
c) x^3 + 27 + (x + 3)(x - 9)
= (x+3)(x^2-3x+9) + (x + 3)(x - 9)
= (x + 3) (x ^2 -3x + 9 + x - 9)
= (x + 3)(x^2 - 2x) = x(x - 2)(x + 3)
Ta có:
(x + 2)(x + 3)(x + 4)(x + 5) - 24
= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24
= (x2 + 5x + 2x + 10)(x2 + 4x + 3x + 12) - 24
= (x2 + 7x + 10)(x2 + 7x + 12) - 24
Đặt x2 + 7x + 10 = k
=> k(k + 2) - 24 = k2 + 2k - 24 = k2 + 6x - 4x - 24
= k(k + 6) - 4(k + 6)
= (k - 4)(k + 6)
=> (x + 2)(x + 3)(x + 4)(x + 5) - 24
= (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)
= (x2 + 7x + 6)(x2 + 7x + 16)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)
Đặt \(x^2+7x+11=t\)thay vào (1) ta được:
\(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:
\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
= 9.[(x^4+2x^2+1)-x^2] - (x^2+x+1)^2
= 9.[(x^2+1)^2-x^2] - (x^2+x+1)^2
= 9.(x^2+x+1).(x^2-x+1)-(x^2+x+1)^2
= (x^2+x+1).(9x^2-9x+9-x^2-x-1)
= (x^2+x+1).(8x^2-10x+8)
= 2.(x^2+x+1).(4x^2--5x+4)
Tk mk nha nếu đúng
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Let \(t=x^2+7x+10\) we have:
\(=t\left(t+2\right)-24=t^2+2t-24\)
\(=\left(t-4\right)\left(t+6\right)=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)-24\)
\(=\left(x^2+3x\right)+2\left(x^2+3x\right)+1-25\)
\(=\left(x^2+3x+1\right)^2-5^2\)
\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)-24\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+6\right)-24\)(1)
Đặt \(x^2+3x+3=t\)thay vào (1) ta được
\(\left(t-3\right)\left(t+3\right)-24\)
\(=t^2-9-24\)
\(=t^2-33\)
\(=\left(t-\sqrt{33}\right)\left(t+\sqrt{33}\right)\)(2)
Thay \(t=x^2+3x+3\)vào (2) ta được :
\(\left(x^2+3x+3-\sqrt{33}\right)\left(x^2+3x+3+\sqrt{33}\right)\)
(x+1)(x+3)(x+5)(x+8)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
=>x2+8x+15=t+8
=>(x2 +8x+7)(x2+8x+15)+15
=t(t+8)+15
=t2+8t+15
=t2+3t+5t+15
=t(t+3)+5(t+3)
=(t+3)(t+5)
=(x2+8x+10)(x2+8x+12)
Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(\Rightarrow A=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\)
\(\Rightarrow A=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t+1\right)\left(t-1\right)\)
\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x^2+2x+6x+12\right)\left(x^2+8x+10\right)\)\(=\left[x\left(x+2\right)+6\left(x+2\right)\right]\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
x3+27+(x+3)(x+9)
= (x+3)(x2-3x+9)+(x+3)(x+9)
= (x+3)(x2-3x+9+x+9)
=(x+3)(x2-2x+18)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\\ =\left(x+3\right)\left(x^2-3x+9+x-9\right)\\ =\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)